40th Vietnamese Mathematical Olympiad 2002 Problems

40th Vietnamese Mathematical Olympiad 2002 Problems

A1.  Solve the following equation: √(4 - 3√(10 - 3x)) = x - 2.

A2.  ABC is an isosceles triangle with AB = AC. O is a variable point on the line BC such that the circle center O radius OA does not have the lines AB or AC as tangents. The lines AB, AC meet the circle again at M, N respectively. Find the locus of the orthocenter of the triangle AMN.

A3.  m < 2001 and n < 2002 are fixed positive integers. A set of distinct real numbers are arranged in an array with 2001 rows and 2002 columns. A number in the array is bad if it is smaller than at least m numbers in the same column and at least n numbers in the same row. What is the smallest possible number of bad numbers in the array?

B1.  If all the roots of the polynomial x³ + a x² + bx + c are real, show that 12ab + 27c ≤ 6a³ + 10(a² - 2b)^3/2. When does equality hold?

B2.  Find all positive integers n for which the equation a + b + c + d = n√(abcd) has a solution in positive integers.

B3.  n is a positive integer. Show that the equation 1/(x - 1) + 1/(2²x - 1) + ... + 1/(n²x - 1) = 1/2 has a unique solution x_n > 1. Show that as n tends to infinity, x_n tends to 4.

Solution

40th VMO 2002

Problem

Solve the following equation: √(4 - 3√(10 - 3x)) = x - 2.

Solution

Answer: x = 3.

Squaring twice we get x⁴ - 8x³ + 16x² + 9x - 90 = 0. Factorising, we get (x + 2)(x - 3)(x² - 7x + 15), so the only real roots are x = -2 and 3. Checking, we find that 3 is indeed a solution of the original equation, but x = -2 is not because we get √(-8) on the lhs.

Thanks to Suat Namli for a similar solution.

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40th VMO 2002

Problem A2 ABC is an isosceles triangle with AB = AC. O is a variable point on the line BC such that the circle center O radius OA does not have the lines AB or AC as tangents. The lines AB, AC meet the circle again at M, N respectively. Find the locus of the orthocenter of the triangle AMN.

 

Solution

 

Let A' be the reflection of A in BC. Then OA = OA', so the circle also passes through A'. ∠MAA' = ∠NAA', so A' is the midpoint of the arc MN. Hence OA' is perpendicular to MN. Let X be midpoint of MN, so X lies on OA'.

Now ∠OA'M = ∠MA'N/2 = (180^o-∠A)/2 = 90^o - ∠A/2, so ∠MOX = 2(90^o-∠OA'M) = ∠A. Hence OX/OA' = OM cos A/OA' = cos A, which is fixed. So the locus of X is line parallel to BC. Let G be centroid of AMN, then AG/AX = 2/3, so G also lies on a line parallel to BC. But H lies on ray OG with GH = 2OG (Euler line), so H also lies on a line parallel to BC.

Given any point H on the line take a line through A' parallel to AH. It meets the line BC at a point O, which is the required point to generate H. (Arguably, we do not generate the two points on the line corresponding to OA perpendicular to AB and AC, because then one of M, N coincides with A and AMN is degenerate.)

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